Consider "gluing" the two men together, so that there are really four slots to consider.

They can sit in seats 1&2, 2&3, 3&4, 4&5.

So we have essentially, four ways the two men (glued together) can sit next to each other, and then 3! ways in which the three women can take up the remaining seats 3. That gives us $4\cdot 3! = 4!$. In addition, the two man can swap positions, so we can "unglue" them, swap them, thus doubling the possibilities. This gives us $2\cdot 4!$ ways in which the men will be sitting together.

Now, there are $5!$ total possible combinations in which five total people can sit in 5 consecutive seats: 5 possible seats in which person one might sit, then 4 remaining seats where person two might sit, down to 3 seats options for person 3, etc until only one remaining seat where the fifth person must sit. This gives us $5\cdot 4\cdot 3 \cdot 2 \cdot 1 = 5!$ possible arrangements in which 5 people can sit.

So our probability $p$ that two men will sit next two each other will be the
(number of arrangements in which two men must be seated together), divided by (the total number** of all possible seating arrangements, without restrictions), giving us: $$p = \dfrac {2\cdot 4!}{5!} = \dfrac 25 = 0.4$$

Of course, we are assuming here that every possible seating arrangement has equal probability: i.e., that the assignment of any one of the five people to any of five consecutive seats is random.

We will choose a pair of seats and put a "Reserved for Men" sign on it. The pair of seats can be chosen in $\binom{5}{2}$ equally likely ways.

Now think of the seats as X X X X X. There are $4$ ways to choose an adjacent pair of seats on which to put the Reserved signs. For the leftmost chosen seat can be any of the first four. Thus the required probability is
$$\frac{4}{\binom{5}{2}}.$$